Why is DeltaG Negative for H2O(l) to H2O(g) at Equilibrium?

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Discussion Overview

The discussion centers around the Gibbs free energy change (ΔG) for the phase transition of water from liquid to gas at 100 degrees Celsius, particularly in the context of equilibrium conditions and vapor pressure. Participants explore the implications of ΔG being negative despite the expectation that it should be zero at equilibrium.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant calculates ΔG for the reaction H2O(l) = H2O(g) at 100°C and questions why it is negative if the reaction is at equilibrium.
  • Another participant asks for the equilibrium vapor pressure of water at 100°C, suggesting a need for clarity on equilibrium conditions.
  • Concerns are raised about the validity of calling the system at equilibrium when the partial pressure of water is only 0.1 bar, which is less than the expected 1 atm.
  • It is noted that the reaction can only be at equilibrium at a specific partial pressure of water, implying that the current conditions do not represent true equilibrium.
  • Participants discuss the notation of the reaction, with one suggesting that using an arrow instead of an equals sign may better represent the reversible nature of the phase transition.
  • Clarification is provided that at pressures below 1 atm, evaporation exceeds condensation, while above 1 atm, condensation exceeds evaporation, indicating dynamic processes at play.

Areas of Agreement / Disagreement

Participants express disagreement regarding the characterization of the system as being at equilibrium, with some arguing it is not due to the low partial pressure of water. There is no consensus on the correct notation for the reaction.

Contextual Notes

The discussion highlights the dependence on specific pressure conditions for equilibrium and the implications of using different notations to represent reversible reactions. The assumptions regarding equilibrium and the definitions of ΔG are not fully resolved.

Kqwert
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Please post this type of questions in HW section using the template.
Calculate deltaG for the reaction:

H2O(l) = H2O(g). 100 degrees celsius, water is clean. P(H2O) = 0.1 bar.

Given that it is an equilibrium, I'd think that deltaG would be zero. But the answer is in fact negative. How is that possible?
 
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What is the equilibrium vapor pressure of water at 100°C?
 
1 bar? So it's really not an equilibrium..?
 
At equilibrium, the partial pressure of water would be 1 atm. Because P(H2O) is less than that, the forward reaction is favored.
 
So it's wrong to call it an equilibrium?
 
The reaction is at equilibrium only at a specific P(H2O).
 
Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?
 
H2O(g) ##\rightleftharpoons## H2O(l) is the correct notation.

The reaction is reversible, with vapor molecules continuously condensing into the liquid phase and liquid molecules continuously evaporating into the gas phase. When P(H2O) is less than 1 atm, the rate of evaporation exceeds the rate of condensation. When P(H2O) is greater than 1 atm, the rate of condensation exceeds the rate of evaporation. Only when P(H2O) = 1 atm will evaporation and condensation occur at the same rate, so there will be no net exchange of material between the liquid and gaseous phases.
 
Kqwert said:
Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?
I think it would have been better if they had used an arrow. ---->
 

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